How to solve $x^2e^{0.4x}-2$?

Question

I was given this equation for homework in an introductory class to calculus:

$f(x)=x^2e^{0.4x}-2$

We were supposed to find the local maximum and minimum, the asymptotes, the inflection points, and then, we were supposed to draw a graph of the function. I did all of this, however, in order to draw this graph, I also need the roots of this equation (or maybe the teacher only wants us to draw it using all the important points except the roots), but I am quite unable to find a way to solve this algebraically. Does anyone know how?

Answer 1

We can rework the equation as

$$(0.2xe^{0.2x})^2=2\ 0.2^2$$ and using the special function Lambert's $W$,

$$0.2x=W(\sqrt 2\ 0.2).$$

https://en.wikipedia.org/wiki/Lambert_W_function

But you are not deemed to know this function.

Answer 2

Consider that you look for the zero's of function $$f(x)=x^2\,e^{2 x/5}-2$$ for which $$f'(x)=\frac{2}{5} e^{2 x/5} x (x+5)\quad \text{and} \quad f''(x)=\frac{2}{5} e^{2 x/5} (2 x^2+20 x+25)$$ The first derivative cancels in particular at $x=0$ $$f(0)=-2 \quad \text{and} \quad f''(0)=+2$$ So, this is a minimum.

To obtain approximation, use Taylor expansions or the infinite series $$f(x)=-2+\sum_{n=2}^\infty \frac{\left(\frac{2}{5}\right)^{n-2}}{(n-2)!} x^n$$ Truncate to some order $$f(x)=-2+\sum_{n=2}^p \frac{\left(\frac{2}{5}\right)^{n-2}}{(n-2)!} x^{n}+O(x^{p+1})$$

For example, using $p=4$, you would face the quartic equation $$2 x^4+10 x^3+25 x^2-50=0$$ which, if you are very patient, can be solved with radicals. Numercially, the two solutions are $x_1=-1.95393$ and $x_2=1.13364$ while the exact solutions, given in terms of Lmabert function, are $x_1=-2.19262$ and $x_2=1.12849$.

We could do better using a high-order iterative method such as Newton, Halley or Householder. For example $$f(x) \sim \frac{21 x^2+20x-50}{2 x^2-10x+25}$$ Solve for zero the numerator to obtain, as better estimates, $x_1=-2.09103$ and $x_2=1.13865$. We could continue the process; I give you below the next quadratic equations to be solved for better and better estimates $$1291 x^2+1340 x-3150=0$$ $$78493 x^2+82980 x-193650=0$$

The last one gives $x_1=-2.18584$ and $x_2=1.12867$.